Problem 58 - Chemistry Matter and Change

Question

Hot-Air Balloons A sample of air occupies 2.50 \(\mathrm{L}\) at a temperature of \(22.0^{\circ} \mathrm{C}\) . What volume will this sample occupy inside a hot-air balloon at a temperature of \(43.0^{\circ} \mathrm{C} ?\) Assume that the pressure inside the balloon remains constant.

Step-by-Step Solution

Answer

The volume will be 2.68 L at 43.0°C.

Step 1: Understanding Charles' Law

Charles' Law states that, at constant pressure, the volume of a gas is directly proportional to its temperature in Kelvin. This is expressed with the equation \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where \( V_1 \) and \( V_2 \) are the initial and final volumes, and \( T_1 \) and \( T_2 \) are the initial and final temperatures, respectively.

Step 2: Convert Temperatures to Kelvin

Before applying Charles' Law, convert the given Celsius temperatures to Kelvin using the conversion formula \( T(K) = T(°C) + 273.15 \). So, the initial temperature \( T_1 = 22.0 + 273.15 = 295.15 \ K \) and the final temperature \( T_2 = 43.0 + 273.15 = 316.15 \ K \).

Step 3: Apply Charles' Law Formula

Using Charles' Law, set up the equation \( \frac{2.50}{295.15} = \frac{V_2}{316.15} \). Cross-multiply to solve for \( V_2 \): \( V_2 = \frac{2.50 \times 316.15}{295.15} \).

Step 4: Calculate Final Volume

Compute \( V_2 \) using the equation from Step 3: \( V_2 = \frac{2.50 \times 316.15}{295.15} = 2.68 \ L \). So, the final volume of the sample at \( 43.0^{\circ} \mathrm{C} \) is 2.68 liters.

Key Concepts

Temperature Conversion

Temperature conversion is vital when working with gas laws like Charles' Law. These laws require temperatures to be in Kelvin (K) rather than Celsius (°C) for accurate calculations. To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. This conversion ensures that all temperatures are measured on an absolute scale, where 0 K is the lowest theoretically possible temperature, known as absolute zero.

• Formula: \[ T(K) = T(°C) + 273.15 \]
• Example: 22.0°C become 295.15 K

Understanding this conversion is critical for applying any gas law as it allows us to work with temperatures that directly relate to the energy of gas molecules.

Volume-Temperature Relationship

The volume-temperature relationship in gases is beautifully described by Charles' Law. This law states that if the pressure of a gas remains constant, the volume of a gas is directly proportional to its temperature in Kelvin. An increase in temperature results in an increase in volume, and vice versa, as the gas molecules gain energy and move more vigorously, occupying more space.

• Charles' Law Equation: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
• Example: Doubling the temperature in Kelvin results in doubling the volume.

Recognizing and utilizing this relationship allows us to predict how a gas will behave under varying temperatures, which is especially useful in real-world applications like hot-air balloons.

Ideal Gas Law

The ideal gas law is a cornerstone of chemistry and physics. While our original exercise focuses on Charles' Law, understanding the ideal gas law provides a more comprehensive view of gas behavior. The law is expressed by the equation \[ PV = nRT \] where:

• \(P\) is the pressure
• \(V\) is the volume
• \(n\) is the number of moles of gas
• \(R\) is the universal gas constant
• \(T\) is the temperature in Kelvin

The ideal gas law combines several simpler laws (including Charles' Law) into one equation, showing how pressure, volume, and temperature are interrelated for a given amount of gas. Though not every situation meets all assumptions of this law, it provides an excellent approximation in many scenarios, particularly under standard conditions.